Let, $a_1=1, b_1=3, c_1=3$ and $d_1=1$
Where
$a_2=a_1+c_1,b_2=b_1+d_1, c_2=a_1+b_1+c_1 $ and $d_2=a_1+d_1$
and we defined
$$a_n=a_{n-1}+c_{n-1}$$ $$b_n=b_{n-1}+d_{n-1}$$ $$c_n=a_{n-1}+b_{n-1}+c_{n-1}$$ $$d_n=a_{n-1}+d_{n-1}$$
then we have this approximate relation of involvement of $\phi$
$$a_n\phi^{3/2}+b_n\phi^{1/2}-c_n\phi\approx d_n$$
Where $\phi={\sqrt{5}+1\over2}$, which is known as the golden ration, related to the well-known Fibonacci numbers.
Apparently $d_n$ it is almost an integer for any values of n. As $n\to \infty$ we see $d_n \to$ an integer
My question is: Can anyone please explain why does this part $a_n\phi^{3/2}+b_n\phi^{1/2}-c_n\phi$ always result in almost an integer values for any n?
$\phi$ it is an irrational number and it is interesting to see $\phi$ turn the above equation into an almost integer result.
Examples of above approximate equation, give
$$\phi^{3/2}+3\phi^{1/2}-3\phi \approx 1$$ $$4\phi^{3/2}+4\phi^{1/2}-7\phi \approx 2$$ $$11\phi^{3/2}+6\phi^{1/2}-15\phi \approx 6$$ $$26\phi^{3/2}+12\phi^{1/2}-32\phi \approx 17$$ $$58\phi^{3/2}+29\phi^{1/2}-70\phi \approx 43$$
$$\left( a_n\phi^{3/2}+b_n\phi^{1/2}-c_n\phi - d_n \right) = -(1 - \phi^{1/2})^{n+2} $$
Since $1 - \varphi^{1/2} \approx -0.272$, its powers rapidly converge to zero.
I found this formula from:
To rigorously prove the identity, one might simply show that stepping the recursion has the same effect as multiplying by $(1 - \phi^{1/2})$.
There was some good fortune that the formula was particularly simple; I expected to have to do more work to find it.
A more systematic approach to obtain the closed form would be to apply the theory of linear difference equations.